Re: real roots

Hi!

Or if we evaluate f(-1) by the remainder theorem the signs in the quotient followed by theremainder strictly alternate in sign: + - + - + - + so there are no negative roots less than orequal to -1. (Lower bound theorem)

Similarly if we evaluate f(1) by the remainder theorem the signs in the quotient followed by the remainder are all positive. So there are no positive roots greater than or equal to 1. (Upper bound theorem)

If |x|<=1 we have x+3>=2. Also the even powered terms cannot be negative. Hence thecomplete sum is >= 2 for all x in [-1,1].

Hence f(x)=0 has no real roots, and in fact is always positive since if there were a value a outsideof [-1,1] for which f(a)<0 then the intermediate value theorem would guarantee a root betweeneither 1 and a if a>1 or between -1 and a if a<-1.

Or more elegantly along the lines that Jack pointed out: Completing the square on x^2+x+3 we get